Proof that the order of any finite $p$-group is a power of $p$

Question

What is the most concise proof that the order of any finite $p$-group is a power of $p$?

Answer 1

Certainly the Theorem of Cauchy would do it - if $q$ is a prime other than $p$, dividing the order of $G$, then there would be an element of order $q$, a contradiction.

Answer 2

From Lagrange's theorem we know only that |G| can be divided by the order of the elements. So, for example, |G| = n * p^k ,where n is an integer not equal to p, and p^k is the order of some element. By Cauchy's Theorem then, there will be an element of order p. As Hekster notes above, this implies that there cannot be a different prime factor q of |G|, because that would lead to a violation of the p-group assumption. Also, if |G| = n * p^k, where n is not a prime or equal to 1, there is a contradiction since n can be written as the product of primes.